Question

A bicycle wheel of radius 0.3m has a rim of mass 1.0kg and 50 spokes, each of mass 0.01 kg. What is its moment of inertia about its axis of rotation?

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf I=105Kg-m^2}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Radius of wheel (R) = 0.3m

• Mass of rim (M) = 0.1kg

• No. of spokes (N) = 50

• Mass of spokes (m) = 0.01kg

• Lenght of spokes (L=R)=0.3m

$\large\underline\pink{\sf To\:Find: }$

• Moment of inertia about its axis of rotation (I) = ?

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$\large\underline{\underline{\sf Moment\:of\:inertia\:of\:rim\:(I_{rim}):}}$

$\large{\boxed{\sf I_{rim}=MR^2}}$

$\large\implies{\sf I_{rim}=0.1×(0.3)^2}$

$\large\implies{\sf I_{rim}=0.09Kg-m^2 }$

$\large\underline{\underline{\sf Moment\:of\:inertia\:of\:spokes(I_{spokes}):}}$

$\large{\boxed{\sf I_{spokes}=N\frac{mL^2}{3}}}$

$\large\implies{\sf I_{spokes}= 50×\frac{0.01×(0.3)^2}{3}}$

$\large\implies{\sf I_{spokes}=0.015Kg-m^2 }$

$\large\underline{\underline{\sf Total\: Moment\:of\: inertia (I):}}$

$\large{\boxed{\sf I=I_{rim}+I_{spokes}}}$

$\large\implies{\sf I=0.09+0.015}$

$\large\implies{\sf I=0.105Kg-m^2 }$

$\large\red{\boxed{\sf I=105Kg-m^2}}$

Hence ,

Total moment of Inertia (I) is ${\sf 0.105Kg-m^2}$

Answer 2

The moment of inertia about its axis of rotation is $0.105\ kg-m^2$.

Explanation:

It is given that,

Radius of the wheel, r = 0.3 m

Mass of the rim, M = 1 kg

Number of spokes, n = 50

Mass of each spokes, m = 0.01 kg

The total moment of inertia about its axis of rotation is equal to the sum of moment of inertia of the rim and the spokes.

The moment of inertia of the rim is,

$I_1=Mr^2$

$I_1=1\times (0.3)^2$

$I_1=0.09\ kg-m^2$

The moment of inertia of the 50 spokes is,

$I_2=N(Mr^2/3)$

$I_2=50\times (0.01\times (0.3)^2/3)$

$I_3=0.015\ kg-m^2$

Total moment of inertia,

$I=I_1+I_2$

$I=0.09+0.015$

$I=0.105\ kg-m^2$

So, the moment of inertia about its axis of rotation is $0.105\ kg-m^2$. Hence, this is the required solution.

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